ASTM052 Extragalactic Astrophysics Notes 4 of 6 (QMUL)

7/30/2019 ASTM052 Extragalactic Astrophysics Notes 4 of 6 (QMUL)

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Department of Physics

ASTM-052: Extragalactic Astrophysics

byPeter Clegg

Note 4. Active Galaxies


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ASTM-052 Extragalactic Astrophysics Note 4

P E Clegg 2001 - i - Version 1.0 (16/10/2001)

Table of ContentsChapter 4 : Active Galaxies ........................................................................................................................ 1

1. Introduction ...................................................................................... ......................................................................... 1

2. Properties of Active Galaxies ................................................................. .................................................................. 12.1 Overview ............................................................................... .............................................................................. 1

2.2 Seyfert Galaxies .............................................................................. .................................................................... 22.3 Quasars and BL Lacertae Objects ....................................................................... ............................................... 22.4 Radio Galaxies ................................................................................ .................................................................... 3

3. Models of Active Galactic Nuclei ........................................................................................... ................................. 43.1 The Source of Energy in AGN ............................................................................ ............................................... 4

3.1.1 Nuclear or Gravitational Power? ................................................................................................... .............. 43.1.2 Accretion Power .................................................................................... ...................................................... 43.1.3 The Eddington Limit ............................................................................. ...................................................... 5

3.2 Characteristic Temperatures ......................................................................... ...................................................... 63.3 Accretion Discs ............................................................................... .................................................................... 7

3.3.1 Structure of the Disc ............................................................................. ....................................................... 73.3.2 Conservation of Mass ....................................................................................... ........................................... 7

3.3.3 Conservation of Angular Momentum ...................................................................... ................................... 83.3.4 Conservation of Energy .............................................................................. ................................................. 83.3.5 Viscosity in the Disc............................................................................. ...................................................... 93.3.6 Spectrum Radiated by Disc ........................................................................................ ............................... 10

4. Models of Radio Sources ............................................................... ......................................................................... 124.1 Synchrotron Radiation ................................................................................. .................................................... 12

4.1.1 The Frequency of Synchrotron Radiation .............................................................................. ................... 124.1.2 The Power emitted by a Relativistic electron .......................................................... ................................. 124.1.3 The Synchrotron Spectrum............................................................. ........................................................... 124.1.4 Lifetime of Relativistic Electrons............................................ .................................................................. 13

4.2 Models of the Radio Lobes................................................... ............................................................................ 134.2.1 The Energy in the Lobes...... ......................................................................... ............................................. 134.2.2 The Supply of Energy to the Lobes......................................... .................................................................. 15

4.2.3 Motion of the Jets and Lobes............................. ........................................................................... ............. 164.2.3.1 Superluminal Velocity in Jets .................................................................................. .......................... 164.2.3.2 Velocity of the Lobes ............................................................................ ............................................. 17

4.2.4 Confinement of the radio Lobes............. ............................................................................. ...................... 184.2.4.1 Diameter-Separation Ratio .................................................................... ............................................. 184.2.4.2 Inertial Confinement ................................................................................... ........................................ 194.2.4.3 Ram-Pressure Confinement.............................................. .................................................................. 194.2.4.4 Thermal Confinement ............................................................................ ............................................. 204.2.4.5 Summary and Conclusions .................................................................... ............................................. 20

4.3 The Particle Energy Spectrum ..................................................................... ..................................................... 204.3.1 The Need for an Acceleration Mechanism.................................................................................... ............ 204.3.2 Stochastic Acceleration ........................................................... .................................................................. 20

4.3.2.1 General Scheme ..................................................................................... ............................................. 204.3.2.2 Elastic Collisions between Particles................................................................................... ................ 20

4.3.2.3 Growth of Energy ................................................................................. .............................................. 214.3.2.4 Loss-rate............................................................................ .................................................................. 224.3.2.5 Resultant Energy Distribution ................................................................................................ ............ 22

4.3.3 Shock Acceleration......................................................................... ........................................................... 224.3.3.1 Energy-Gain....................... .................................................................................................. ............... 224.3.3.2 Loss-rate............................................................................ .................................................................. 234.3.3.3 Resultant Energy Distribution ................................................................................................ ............ 23

Bibliography for Chapter 4 ............................................................................. ............................................................ 23


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CHAPTER 4: ACTIVE GALAXIES

1. Introduction

The number of known types of galaxies showingunusual activity has grown over the years to the point atwhich the activity itself can hardly be considered

unusual, even if its origin is not well understood.Moreover, similar sorts of activity, albeit on a smaller scale, are seen in objects usually considered to benormal. Many astronomers are therefore coming tothe view that we are seeing different mixtures, ondifferent scales, of one or two basic phenomena.

In this section, I shall first identify the various types of active galaxy and discuss their properties. I shall thenreview the whole phenomenon.

2. Properties of Active Galaxies

2.1 Overview

Active galaxies are frequently described as showingviolent or energetic activity. Perhaps a more usefuldefinition is galaxies displaying phenomena that cannot

be ascribed to normal stellar processes, although this begs the question of what constitutes a normal stellar process. It could be, for example, that the activity seenin some nuclei is the result of a violent burst of star-formation. In all cases, the activity seems to beconnected with the nucleus of the galaxy, whence theterm active galactic nucleus (AGN). In general, AGNshow some combination of the following features:

1. They have bright, star-like nucleus. In a short-exposure image of the galaxy, only the point-likenucleus is apparent. On longer exposure, the

nucleus saturates the detector photographic plateor CCD and the rest of the galaxy becomesapparent, although for quasars , this is only seenwith great difficulty.

2. The nucleus radiates over a very wide range of wavelengths, from radio (in some cases) to X-rays.This is unlike stars, which radiate mostly at opticalwavelengths, and normal galaxies, which aredominated by starlight and the infrared emissionfrom cool interstellar dust.

3. The spectra of AGN are unlike those of normalgalaxies, being very blue and having no absorptionlines characteristic of starlight, but having strongemission lines.

4. The spectral energy distributions (SED), defined by

( ) S =:SED , (2.1)

of AGN are quite unlike those of normal galaxies.For most AGN, the SED is roughly constant butfollows a power law from the hard X-ray region of the spectrum to the far-infrared (far-IR), theradiation being unpolarised. In radio-loud AGN,this continues out to the radio region. In blazars ,

the SED has a smooth broad hump and is strongly polarised.

5. The SEDs typically have two bumps on top of the power-law.

The blue bump is an excess, rising from the opticalto the ultraviolet (UV). There is some evidence thatthis bump is seen descending again in the soft X-ray region in which case it presumably peaks in theextreme ultraviolet (XUV), at ~ 10 16 Hz; thiscorresponds to a temperature of ~ 300,000 K. The

peak cannot be observed directly because theGalactic interstellar medium is opaque in thisregion of the spectrum.

The X-ray bump occurs in the hard X-ray regionwith ( ) ( )7.0~ S [so that S ( ) is rising withfrequency] and is still rising at E > 20 keV,corresponding to a temperature of ~ 10 8 K.

6. The strong emission lines are much too highlyexcited to be the result of ionisation from OB stars

and are presumably ionised by the photons from the blue bump. The lines can be very broad, presumably the result of Doppler broadening withvelocities ~ 10,000 km s -1.

7. Many radio sources have double radio lobes ,symmetrical blobs on either side of the centralgalaxy. These are very large structures, tens of kiloparsecs in size and separated from the centralgalaxy by ~ 50 kpc to 1 Mpc. The spectrum of radioemission from these lobes is a power law of theform ( ) ( )7.0~ S . There is often a bridge of radio emission between the lobes and the centralgalaxy.

8. Many radio galaxies have relativistic jets of material pointing at the lobes. These jets seem to bemade up of blobs of material that appear to bemoving away from the galaxy faster than light!Although we shall see that this is an opticalillusion, it does mean that the material in the jetsmust be travelling at velocities close to that of light.

9. The brightness of AGN can vary very rapidly. TheX-rays from NGC 4051, for example, change by afactor of 2 in 30 minutes! The variability tends todepend upon wavelength:

X-ray Hours to days

Optical Weeks to yearsFar-infrared Invariable

10. Since an object's output cannot change coherentlyfaster than the time it takes light to cross it, this

puts limits on the sizes of the emitting regions andsuggests enormous powers are being generated invery small regions, sometimes smaller than thesolar system (see below).

11. The total power of AGN is hard to estimate becauseof the great spread of wavelengths and also becauseof the variability. The weakest known AGN are in


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nearby galaxies with luminosities L ~ 10 33 W,whilst the most powerful distant quasars have

L ~ 10 40 W. You should compare this with our Galaxy, which has L ~ 10 10 Lsun or ~ 10

33 W. In themost luminous cases, the nucleus can outshine thewhole of the rest of the galaxy!

Before trying explain these observations, let us look briefly at the various types of active galaxy; a fuller discussion is given in [1]

2.2 Seyfert Galaxies

Carl Seyfert first noticed a class of spiral galaxy withvery bright nuclei in 1943. These were the first AGN to

be recognised and represent the mildest form of activity.As well a dominating the light from the rest of thegalaxy, the output from Seyfert nuclei can vary in lessthan a year. This sets an upper limit to the size of theregion responsible for the emission.

r

t

t=r/c

Figure 4-1. Limits on size of variable objects.

The upper part of Figure 4-1 shows schematically a point source whose luminosity increases and decreaseson a time-scale t . The lower part of the figures showsan extended source, each part of which brightens andfades on the same time-scale t . The overall

brightening and fading of the source, as seen by anobserver, occurs over a longer time-scale t because theobserved change in luminosity of the more distant partsof the source arrives later than that of the nearer parts.In general, we deduce that in an object whoseluminosity varies on a time-scale t , size of the regionresponsible for the is emission is cannot be larger thanr , where

t cr


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( ) constantloglog += S , (2.4)

where S o is the flux-density at the frequency o and isa constant called the spectral index 2. Equation (2.4)shows that the spectrum is a straight line on a log-log

plot of flux-density against frequency.

3. Models of Active Galactic Nuclei3.1 The Source of Energy in AGN

3.1.1 NUCLEAR OR GRAVITATIONAL POWER ?

We have to explain luminosities of at least 10 39 W(10 13 Lsun), generated within regions only 10

13 m or lessacross and lasting for some 10 8 years. In other words,we need to account for the generation of about 10 54 J of energy. [A related, but smaller, problem is theexplanation of the 10 52 J or so found in the lobes of radio galaxies.] Relativity tells us that, in order to

produce a luminosity L, we need to convert mass intoenergy at the rate m& given by

2c

Ldt dm

m =& ; (3.1)

the minus sign occurs because the mass is decreasingwith time. From equation (3.1) we find that we need toconvert about a tenth of a solar mass a year to produce1039 W. At first sight, this seems modest: a quasar would use up only 10 5 solar masses perhaps 10 -6 of themass of a galaxy in 10 8 years. What we have not takeninto account, however, is the efficiency with which massmay be converted into energy. Let us define theefficiency of an energy-generating process as the ratioof the energy E produced to the total rest-mass energyof the material generating the power:

2:

mc

E = . (3.2)

Taking this efficiency into account, equation (3.1) becomes

2

1

c

Lm

=& . (3.3)

Suppose we try to produce 10 39 W of luminosity by asnuclear burning within stars 3. The most efficient nuclear

2Note that some authors define the spectral index by:

( )

=

ooS S

.

This is because most of the early observations were of sources whosespectral index was, by this definition, positive. The slope of log S ( )against log can have either sign, however, and it is becoming moreusual but not universal to use the form I have chosen.3Even if we were to do this, we should still have to explain how toconvert ordinary stellar radiation into relativistic particles needed toexplain the observed spectra see later.

process is the conversion of hydrogen into helium, with = 0.7%. The rate of star-formation that is the rate m& at which interstellar material is converted into stars isgiven by equation (3.3) as

( ) .yM25skg106.1

skg103007.0

10

1-sun

1-24

1-28

39

=

m&(3.4)

This is a very high rate: in 10 8 y, a galaxy wouldconsume some 10 9 Msun , or perhaps 0.1% of its mass.And, if we are to explain the variability of AGN, weneed all this mass to be contained within a region notmore than 10 13 m across. The binding energy of massm contained within a region R in radius is given by

RGm 2

~ (3.5)

and is about 10 55 J for the above figures. This bindingenergy, which could in principle be released bygravitational contraction, is about an order of magnitudemore than we are trying to produce by nuclear means!This suggests that the release of gravitational energy isthe more likely source of power in AGN, although weneed to look at the efficiency of this process as well.

3.1.2 ACCRETION POWER

Let us see if we can find a more efficient process.Consider the decrease in gravitational potential energy of a body of mass m when it falls to withindistance r of another body of mass M :

r

GMm

= . (3.6)

If the body is in free-fall, however, this decrease in potential energy will simply go into increasing thekinetic energy of the body. We have to find a way of converting the potential energy into radiation. The usualmodel of this process is to consider the body to begaseous material in an accretion disc and which isslowly spiralling into the central mass 4. In order tospiral, the material must be slowly losing angular momentum and this is envisaged as arising fromviscosity in the disc mutual friction between thematerial successive turns of the spiral 5. This viscosityheats the disc, converting the kinetic energy of the

orbiting material into internal energy of the moleculesand atoms of the disc. Finally, this heat is radiated away

by photons. Let us look at this process morequantitatively.

4 This is almost certain t o happen anyway because the in-fallingmaterial will have angular momentum with respect to the centralobject that it will have to get rid of. 5It has to be said that no satisfactory viscous mechanism has beenfound!


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Since the material is in quasi-stationary orbit about thecentral mass, the kinetic energy T (r ) and potentialenergy (r ) of material at distance r from the centremust obey the virial theorem:

( ) ( ) 02 =+ r r T . (3.7)

The total energy of the system that is the sum of thekinetic energy, the potential energy and any energy R(r )that has been radiated by the time the material hasreached r must be conserved, so that

( ) ( ) ( )0=++ r Rr r T , (3.8)

where I have taken the potential energy of the materialat infinite distance from the central mass to be zero.Eliminating T from equations (3.7) and (3.8), we get

( ) ( )

( ) ( ).

2

1

2

1

;21

21

r

GMmr r T

r GMm

r r R

==

==(3.9)

Only half the potential energy is, therefore, available to be converted into radiation.

It appears at first sight as if we could extract infiniteenergy from this process by allowing r to go to zero.There is, however, a natural limit to how close one canapproach to an object of mass M . General relativity tellsus that, if a body becomes smaller that its Schwarzschild radius r S, given by

2S2

:c

GM r = , (3.10)

it will collapse to form a black hole from which nothingcan escape. Putting in numbers, we get

( )

=

sun

9S 1076.2hourslight M

M r . (3.11)

Hence the Schwarzschild radius represents an absoluteminimum to the value of r occurring in equation (3.9)and then only if the body of mass M is, indeed, a black hole. Because we get maximum efficiency from a black hole, let us shall assume that this is the case. Generalrelativity also says that there is no stable orbit closer toa black hole than 3 r S. Material closer than this plungesstraight into the hole without having time to get rid of its remaining gravitational energy 6. In practice,therefore, the maximum radiant energy Rmax that I canextract from letting matter of mass m fall into a black hole is given by

2

smax 12

132

1mc

r GMm

R == (3.12)

6This is a slight over-simplification but it serves our purpose.

with a corresponding maximum luminosity L given by

2max 12

1cm

dt dE

L &= . (3.13)

The minimum mass-inflow rate m& needed to produce aluminosity L is therefore given by

212

c

Lm =& (3.14)

Comparing equation (3.14) with equation (3.3), we seethat the efficiency of gravitational accretion is 1/12 or about 8%, which is an order of magnitude more efficientthan nuclear processes. This means that only about2 solar masses a year are needed to fuel the observedluminosities. Actually, we can expect the efficiency to

be even higher than this. I have dealt only with a non-rotating black hole. In practice, any black hole formedfrom the collapse of material to the centre of a galaxy islikely to be rotating because the material from which it

formed would have had angular momentum about thecentre of the galaxy. The maximum efficiency of extracting potential energy from a rotating black holecan be shown to be about 40%.

Note that the luminosity predicted by this modeldepends only on the mass-infall rate and is independentof the mass of the black hole itself. The observedvariability, however, limits the size of the region andhence the Schwarzschild radius. Variability on the timescale of an hour demands a black hole of at least109 Msun .

3.1.3 THE EDDINGTON LIMIT

It is easy to show that there is a limiting accretion-

luminosity for a body of a given mass. Out-flowing photons exert a force on the in-flowing matter and, if the flux of photons is large enough, this force willexceed the gravitational attraction of the central mass.The situation is illustrated in Figure 4-3.

r

F in

Fout

electron

proton

Figure 4-3. Origin of the Eddington Limit

The details are as follows. The general relation betweenthe energy E and momentum p~ of a relativistic particleof mass m is given by

42222 ~ cmc p E += . (3.15)


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Since photons are massless, each carries momentum p~ given by

c E p =~ (3.16)

and the total momentum P~

carried per unit time by all

the photons emitted by a source of luminosity L is given by

c L

P =~ . (3.17)

The pressure p exerted by these photons is, bydefinition, the momentum flux, that is the rate flow of momentum per unit area of surface perpendicular to theflow of radiation. At a distance r from the source,therefore,

( ) cr

Lr p

=

24 . (3.18)

This pressure acts upon the Thomson cross-sections of the (ionised) in-falling particles, mainly protons andelectrons. For a particle of charge e and mass m,

( )2

2o

2

432

,

=

mc

eme

(3.19)

It is clear from equation (3.19) that, because of the ratioof their masses, the Thomson cross-section of electronsis some six orders of magnitude greater than that of

protons. The main outward force F out(r ) is therefore thatexerted by the photons on the electrons and is given by

( ) ( )cr

Lr pr F

2T

Tout4

== , (3.20)

where T is the Thomson cross-section of the electron.Because the electrons in the plasma are coupled to the

protons via electromagnetic forces, the outward forcegiven by equation (3.20) is transferred to the protons aswell.

The inward gravitational force F in(r ) on a particle of mass m is given as usual by

( )2

inr

GMmr F = . (3.21)

Equation (3.21) shows that the gravitational force on the proton is three orders of magnitude higher than that onthe electron so that the main inward force is given by

( )2

Pin

r

GMmr F = , (3.22)

where m p is the protons mass. Again, though, this forceis transferred to the electrons electromagnetically.

For accretion to take place, we need inward force given by equation (3.22) to be greater than the outward forcegiven by equation (3.20):

cr

L

r

GMm2T

2P

4

. (3.23)

This means that, if a source is to derive its luminosityfrom accretion, its luminosity L must be less than the

Eddington luminosity , LEddington :

T

pEddington 4:

cGMm

L L = . (3.24)

Putting in numbers, we get

( )

=

sun

31Eddington 1026.1W M

M L . (3.25)

Relation (3.24) can be re-arranged to give a lower limit

on the mass of a source with a given observed velocity L:

cGm L

M M p

Tmin 4

1

= . (3.26)

Again putting in numbers, we get

( ) ( )W109.7M 32sunmin L M = . (3.27)

Equation (3.27) shows that we need a black hole of atleast 10 8 solar masses to give an accretion luminosity of 1039 W.

3.2 Characteristic TemperaturesLet us make some rough estimates of the temperaturesinvolved in accretion radiation. First, consider thatcoming from the inner edge of the accretion disc atradius r min . If we assume that the radiation is blackbodywith effective temperature T eff , then the luminosity L isgiven by

4eff

2min~ T r L , (3.28)

where is the Stefan-Boltzmann constant. If we assumethat the disc is radiating at the Eddington limit and thatr min is the radius of the last stable orbit, then we havefrom equations (3.28), (3.24) and (3.10),

4/1

T

5 p

eff 9~

GM

cmT . (3.29)

Putting in numbers, we have

4/1

sun

7eff 104~

M

M T . (3.30)


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For a 10 8 solar mass black hole, we find an effectivetemperature of about 10 5 K. Using the relation

kT hc

~

, (3.31)

we see that this radiation peaks at around 150 nm, or inthe UV to soft X-ray region of the spectrum where the

big blue bump lies. Moreover, from equation (3.10), wesee that the size of this region is about a light-hour across, so we should not be surprised at variations onthe time-scale of hours.

On the other hand, think of protons that fall straightfrom infinity to r min and are then thermalised. Thegain in T in kinetic energy of each proton is given,from equation (3.9), by

2 p

S

p

61

3cm

r

GMmT == . (3.32)

If this energy is now thermalised to a temperature T therm ,we have

therm23

kT T = (3.33)

so that

k

cmT

2 p

therm 91= (3.34)

or, numerically

K 102.1 12therm =T .7 (3.35)

From equation (3.31), we find that the energy of the photons corresponding to this temperature is about100 MeV, many times the energy (1 MeV) required for the production of electron-positron pairs. From this, wemay tentatively conclude that:

the big blue bump arises from optically thick thermalradiation from the inner parts of the accretion disc;

the X-rays originate from a region of electron- positron plasma.

3.3 Accretion Discs

3.3.1 STRUCTURE OF THE DISC Figure 4-4 shows schematically an accretion disc of density (r) encircling a black hole and though whichmass is spiralling down at a rate m& 8. The upper part of the figure shows the plan view and the lower part a

7 Note that these results are independent of the mass of the black hole.8If the disc is to be in a steady state, that is of there is to be no build upof matter anywhere in it, the mass infall rate must be independent of radius.

section. Consider a thin annulus of the disc of width r at r . Define the surface density (r ) of the disc at r by

( ) ( )( )

( )r d r r

r h

r h

= +

2/

2/

: , (3.36)

where h(r ) is the thickness of the disc at r .

rr + r

h(r)

(r)

Figure 4-4. Accretion disc.

3.3.2 CONSERVATION OF MASS

Consider an annulus of width r at r . The rate of build-up of mass ( ) t r m within this annulus is given by

( ) ( )[ ]r r r t t

r m

2 . (3.37)

This build-up must be provided by the net flow of material

netm& into the annulus:

( )outinnet mmmt

r m &&& ==

, (3.38)

where inm& is the rate of flow of mass into the annulus

from the outer regions and outm& is rate of flow of massfrom the annulus into the inner regions. Now

( ) ( ) ( )r vr r r mm r = 2out && . (3.39)

and

( )( ) ( ) ( ),2

in

r r vr r r r

r r mm

r

+++=+ &&

(3.40)

where vr (r ) is the radial component of velocity of thematerial in the disc (positive in the direction of increasing r ). Hence,

( ) ( )[ ]r vr r r

r m r = 2net& (3.41)

to first order in r . From equations (3.37), (3.38) and(3.41), we have


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( )[ ] ( ) ( )[ ]r vr r r

r r t

r r r =

22 (3.42)

or

( ) ( ) ( )[ ] 01 =+

r vr r r r t

r r

. (3.43)

In the steady state, ( ) t r must be zero so that

( ) ( ) constant=r vr r r . (3.44)

Comparing equations (3.44) and (3.39), we see that

( ) ( ) ( ) mr vr r r m r && === constant2 , (3.45)

independent of r . That is, the radial mass flow isindependent of radius, as we expected.

3.3.3 CONSERVATION OF A NGULAR MOMENTUM

Now consider the transfer of angular momentum across

the annulus. The rate ( )r L flowmass& at which clockwiseangular momentum is flowing out to the inner regionsof the disc is given by

( ) ( ) ( ),flowmass r vr r mr L = && (3.46)

where v (r ) is the azimuthal velocity of the material at r .taken to be positive clockwise. Using similar argumentsto those above, we find that the rate of build-up

( ) t r L of angular momentum in the annulus, resultingfrom the mass flow, is given by

( )( )[ ]r r rvr mt

r L

=

&flowmass

, (3.47)

where I have used the constancy of mass-flow, equation(3.45)

There is, however, another source of angular momentum the action of any torque (r ) acting in thedisc: I shall derive an expression for (r ) later. If wetake positive (r ) to be clockwise acting on materialinterior to r , then the net clockwise torque (r ) actingon the annulus is given by

( ) ( ) ( )( )

r r

r

r r r r

+=(3.48)

so that, from equations (3.47) and (3.48)

( ) ( ) ( )

( )[ ] ( )r r

r rvr

m

r r

t r L

t r L

+

=

+

&

flowmass

. (3.49)

In the steady state, ( ) t r L must be zero so

( )[ ] ( )r r

r rvr

m =

& . (3.50)

If we integrate equation (3.50) between r lso, the radius of the last stable orbit, and r , we get

( )[ ]( )

=

r

r

r

r r d r

r r d r vr r m

lsolso

& , (3.51)

or

( ) ( ) ( ) ( )[ ]( ),

lsolsolso

r

r r r vr r rvm

== & (3.52)

where I have assumed that the torque at the last stableorbit is zero, because there is nothing for it to act on.

3.3.4 CONSERVATION OF E NERGY

Finally, let us consider the conservation of energy. We

see that the rate ( )r T &

at which kinetic energy is flowingout of the annulus into the inner regions of the disc isgiven by

( ) ( )r vmr T 221

= && . (3.53)

Proceeding along similar lines to the above, we find thatthe net rate of flow ( ) t r T of kinetic energy into theannulus is given by

( ) ( )[ ]r r vr

mt r T

2

21

=

& . (3.54)

We must also take account of the flow of potentialenergy V . The rate ( )r V & at which potential energy isflowing out of the annulus to inner regions is given by

( ) ( ) ( )

,

2

r mGM

r GM

r vr r r V r

&

&

=

=

(3.55)

where I have used equation (3.45) The net rate of increase ( ) t r V in the annulus is therefore given by

( )r r r mGM t

r V

= 1

& . (3.56)

Finally, the differential torque does work on theannulus. This net rate of working ( ) t r W is given by


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( ) ( ) ( ) ( ) ( )

( ) ( )[ ] ( )( )

( )

( )( ) ,1 2lsolso r r v

r v

r v

r

r

r

m

r r

r vr

r r r r

r

r r r r r r t r W

=

=

++=

&

(3.57)

where (r ) is the angular velocity of the material at r and where I have used the momentum conservationequation (3.52) to substitute for (r ). Hence, the overallrate ( ) t r E of build-up of total energy in the annulusis given by

( ) ( ) ( ) ( )

( )

( )( ) ( )

( )( ) ( ) r

r GM

r vr v

r v

r r

r m

r

r vr v

r v

r r

r GM

r v

r m

t r W

t r V

t r T

t r E

+

=

=

+

+

=

2lsolso

2lso

lso

2

21

1

21

&

& . (3.58)

If the material is in almost circular Keplerian orbitsabout the black hole ( vr and mean-free-

path , superimposed on its circular and radial motions.Consider a blob of material A in Figure 4-5, movingfrom the initial position shown to a new position atsmaller radius. Then the mass-transfer rate chaosm&

produced by such movement is given by

( ) vr r m = 2chaos& (3.64)

On average, such a blob will transfer angular momentum corresponding to material at r + across thesurface of at radius r . Similarly, blob B will transfer

angular momentum corresponding to material at r - across the same surface. This net transfer of angular momentum will exert a torque on the disc at r .

r

r + r -

A

B

Figure 4-5. Origin of Viscosity.

Material at r has circular velocity r (r ). If the discrotated as a rigid body, so that (r ) were constant, thenthe material at ( r + /2) would have circular velocity(r + /2)(r ). In fact, the material at ( r + /2), where Aoriginates, has circular velocity ( r + /2)(r + /2). The

9 I am assuming that the disc is optically thick.


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relative velocity of material at ( r + /2) with respect tothat at r is therefore given by

( )

( )

( ) ( ) ( )

( ),21

22

22

222

r r

r r r r

r r r

r r r r

+

+

+

+=

+

+

+

(3.65)

where I have used a Taylor expansion and kept onlyfirst-order terms in . The rate of transfer of angular momentum ( )r Lin& across r by material such as A istherefore given by

( ) ( )

= r r r mr L 21

chaosin&& . (3.66)

Similarly, the rate of transfer of angular momentum( )r Lout& across r by material such as B is given by

( ) ( ) ( )

= r r r mr L 21

chaosout&& . (3.67)

The net rate of transfer of angular momentum inward( )r L& at r is therefore given by

( ) ( ) ( )( )( ) ( ),2 3

chaos2

outin

r vr r

mr r

r Lr Lr L

==

=

&

&&&

(3.68)

where I have used equation (3.64). But the net rate of transfer of angular momentum across a surface is equalto the torque exerted on that surface. The torque (r )exerted, in the direction of rotation, by the outer material on the inner material at r is therefore given by

( ) ( ) ( )r vr r r = 32 , (3.69)

where

( ) vr =: (3.70)

is the coefficient of viscosity. Note that, if (r ) isactually constant, independent of r , then (r ) is zero: weonly get a torque of there is shear motion in the disc.

3.3.6 SPECTRUM R ADIATED BY DISC

Let us assume that the disc radiates as a black body sothat

( ) ( )r T r l 4 = , (3.71)

where T (r ) is the temperature of the disc at r and is theStefan-Boltzmann constant. From equations (3.71) and(3.63), we have

( )

4/12/1

2

4/1

3

4/12/1lso

4/1

3

61

83

183

=

=

r c

GM

r

mGM

r r

r

mGM r T

&

&

. (3.72)

The spectrum I ( ,T ) of black-body radiation attemperature T is given by

( )1

12,

2

3

=

kT hec

hT I

(3.73)

so that the spectral luminosity l( ,r ) per unit area of theaccretion disc is given by 10

( ) ( ) 112,

2

3

=

r kT hechr l

, (3.74)

where T (r ) is given by equation (3.72). The totalspectral luminosity L( ) of the disc is therefore given by

( ) ( )

( )

( )

=

=

max

lso

max

lso

max

lso

18

1

124

2,2

2

32

2

3

r

r

r kT h

r

r

r kT h

r

r

erdr

ch

rdr ec

h

rdr r l L

, (3.75)

where r max is the outer radius of the disc and the firstfactor of two recognises that the disc has two faces.

Although the integration of equation (4.26) has to bedone numerically, we can get a good idea of what thisspectrum looks like by making some approximations. Atlow frequencies, we have

.1;1

1


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( ) ( )

( )

,

,85

32

1

88

8

2

24/1

maxlso8

42

4/12/1lso4/1

4/1

2

22

2

22

max

lso

max

lso

=

r r J c

k mGM

dr r

r r

mGM k

c

rdr r T k c

L

r

r

r

r

&

&

(3.78)

where, J (r lso ,r max ) is the integral in the third line of (3.78). At low frequencies, therefore, the spectrum is a

power law.

At high frequencies, we have

.1;1

1 >>

kT h

ee

kT hkT h

(3.79)

Differentiation of equation (3.72) shows that themaximum temperature occurs at r T max given by

lsolso

2

max 8.134

r r r T =

= . (3.80)

For frequencies such that

maxT r kT h >> , (3.81)

we have

( ) ( )

( )

( ),4

8

8

max

max

lso

max

max

lso

2max2

32

2

32

2

32

r kT hv

r

r

r kT hv

r

r

r kT hv

er c

h

rdr ec

h

rdr ec

h L

, (3.82)

where, in the second approximation, I have replaced theexponential term with its maximum value and, in thethird, have assumed that r max >> r lso . At highfrequencies, the spectrum falls off exponentially withfrequency.

What about in between? The blackbody function (3.73) peaks at frequency given by

hkT

~ (3.83)

and the total luminosity per unit area is T 4 equation(3.71). We can approximate l( ,T ) very crudely as

( ) ( ) ( )

h

r kT r T r l 4, , (3.84)

( x) is the Dirac delta function. Substituting fromequation (3.84) into the first line of equation (3.75), weget

( ) ( ) ( )

( )

=

=

max

lso

max

lso

4

4

84

4

r

r

r

r

rdr h

kT r T

mGM

rdr h

r kT r T L

&

(3.85)

The integral could be evaluated by changing thevariable of integration from r to T using equation (3.72).Unfortunately, the term

4/12/1lso1

r r

complicates this considerable. Since we are onlylooking for an approximate behaviour, I shall ignore thisterm. Then

( )( )

( )

3/1

3/13/13/5

3/13/5

8316

84

lso

max

=

=

k hmGM

dT h

kT T

mGM L

r T

r T

&

&

(3.86)

so that L( ) is proportional to 1/3 in the intermediateregion. The overall spectrum is sketched in Figure 4.6.

How does this fit with observation? It is not a badrepresentation of the blue bump in AGN spectraalthough 0 would be better than 1/3 Nevertheless,especially in view of the approximations involved, it isnot a bad attempt.

ln L

ln

2

1/3 e -h /kT

Figure 4.6. Overall spectrum of accretion disc.


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4. Models of Radio Sources

4.1 Synchrotron Radiation

4.1.1 THE FREQUENCY OF SYNCHROTRON R ADIATION

A power-law spectrum of the form given in equation(2.3) is consistent with the synchrotron radiation emitted by a collection of charged particles, of variousenergies, moving with relativistic velocities in amagnetic field. Synchrotron emission is also consistentwith the observed polarisation of the radiation. I shallgive a simplified treatment of such radiation.

An electron moving perpendicularly to a magnetic fieldof induction B describes a circular orbit with thecyclotron frequency c given by

ec 2

1meB

= , (4.1)

where me is the rest-mass of the electron. Notice that c is independent of the velocity (or energy) of theelectron. Because circular motion involves acceleration,the electron will emit electromagnetic radiation withfrequency c. If the electron is moving relativisticallywith velocity u however, it emits radiation over a broad

band of frequencies peaked at given by

2

ec

2

41

2

meB== , (4.2)

where the Lorentz factor is given by

.:

;1

1

1

1222

cu

cu

=

=

=

(4.3)

Using the fact that the energy E of the relativisticelectron is given by

2e cm E = , (4.4)

we get

2243

e41

aBE BE cm

e =

=

, (4.5)

where

=

43e4

1:

cm

ea

. (4.6)

Putting in numerical values, we get

( ) ( ) ( )JT101.2Hz 236 E B= . (4.7)

4.1.2 THE POWER EMITTED BY A R ELATIVISTICELECTRON

The power P( E) radiated by an electron of energy E isgiven by

( ) ( )

22254

e

4

o

22e

24

o

614

1

3

2

E Bcm

ecm

Be E P

=

=

(4.8)

where o is the permittivity of free space and where thesecond line come from equation (3.39).

In any collection if electrons in space, there must be anequal number of protons to preserve charge neutrality,and these protons will also radiate in a magnetic field.Equation (4.8) shows, however, that the power radiated

by a particle in inversely proportional to the fourth power of its mass. Even if the protons are movingrelativistically with the same energy as the electrons,therefore, the power they radiate will be relativelynegligible because they are some two thousand timesmore massive 11 .

In all the cases we shall deal with, the electrons areultra-relativistic so that

1 . (4.9)

We can therefore re-write equation (3.63) as

( ) 22 E bB E P , (4.10)

where

=

54e

4

o61

cm

eb

. (4.11)


( ) ( ) ( )JT1042W 2212 E B.P = . (4.12)

4.1.3 THE SYNCHROTRON SPECTRUM

Suppose that the number N ( E )dE of relativistic electronswith energies between E and E + dE is given by

( )


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( ) ( ) ( ),22oo dE E bB E N

E PdE E N E dL p p =

=(4.14)

where I have used equation (3.73). We can use equation(3.71) to eliminate the energy E from this equation:

( )( )

( ) .21 2

12

1

2/3oo

2/12/2

2oo

d Ba

b E N

aBd

aBbB E N dL

p p

p p

p

p

+

=

=(4.15)

This spectrum has the same form as that given inequation (2.3) provided that

21 p= . (4.16)

Local cosmic rays have an energy spectrum of the form(4.13) with p ~ 2.5, which would give ~ 0.75. This isremarkably close to the commonly observed value of 0.7. We may therefore feel that we have explained interms of cosmic rays in the radio sources. We haveyet to explain, though, why p for cosmic rays shouldhave this value! I shall return to this point later.

4.1.4 LIFETIME OF R ELATIVISTIC ELECTRONS

Because the electron is radiating power P ( E ), it is losingenergy at this rate so

( ) E Pdt dE = . (4.17)

Substituting for P ( E ) from equation (3.58) we find

22 E bBdt dE = . (4.18)

Hence, an electron with energy E loses energy E in atime t given by

22 E bB

E t

= . (4.19)

Crudely, we can put E equal to E and obtain acharacteristic time ( E ) for an electron to lose itsenergy, where

( ) E bB

E 2

1= . (4.20)

Notice that the more energetic the electron, the morerapidly it loses its energy. ( E ) can be considered as thelifetime for radiation from an electron of energy E .

[More formally, we can integrate equation (4.18):

t bB E E d

E

E i

2=

, (4.21)

where E i is the initial velocity of the electron at t = 0.Carrying out the integration on the left-hand side, wefind that

t E bB

E E

i

i21+

= . (4.22)

Even if the electron were to start with infinite energy, itwould have energy E after a finite time t ( E ) given by

( ) E bB

E t 2

1= . (4.23)]

We can use equation (3.71) to eliminate the energy E from (3.41) and obtain the characteristic lifetime ( ) of an electron radiating at frequency :

( ) 2/12/32/1

= B

ba

. (4.24)

Substituting numerical values, we get

( ) ( ) ( ) 2/12/35 HzT106s = B . (4.25)

4.2 Models of the Radio Lobes

4.2.1 THE E NERGY IN THE LOBES

In this section, I shall assume that all electrons have thesame energy E o. That is, I shall replace the true energyspectrum given by equation (3.59) by the simplifiedspectrum

( ) ( )dE E E N dE E N oo = . (4.26)

This will make treatment much easier without changingthe overall conclusions. Suppose the source has totalluminosity L. From equations (4.26) and (3.78), we have

( ) ( )( )

( ).oo

2o

2oo

22o

E P N

E bB N dE E E E bB N

E PdE E N L

===

(4.27)

The total energy E electrons contained in the electrons isgiven by

( )( ) .oooo

electrons

E N dE E E E N

E dE E N E

===

(4.28)

From equations (4.8) and (4.13) we get


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( ) ( )

,2/1o2/3

2/1

o

o

oo

ooelectrons

=

==

LBb

a

L E P

E L

E P N

E N E

(4.29)

where I have used equation (4.5) to eliminate E o. Puttingnumerical values into, we obtain

( ) ( ) ( ) ( )2/1o2/35electrons HzTW106J = B L E .(4.30)

Equation (3.41) takes account only of the energy in theelectrons. If the material is to be electrically neutral, theelectrons must be accompanied by an equal number of

protons, which will also have energy. To allow for this,let us say that the energy E protons in the protons is relatedto E electrons by

electrons protons KE E = . (4.31)

The total energy E particles in particles is then given by

( )

( ) .1

1

2/1o

2/32/1

electrons

protonselectrons particles

+=

+=+

LBb

aK

E K

E E E

(4.32)

We have no way of determining K directly. We mayhave a clue in the cosmic ray flux measured near theEarth in which the protons carry 100 times as mushenergy as the electrons. In the absence of better information, therefore, let us take K to be 100. Then it iseasy to show that the total particle energy in the radio

lobes of Cygnus A, for example, is equivalent to therest-mass of about 10 5 M sun .

A magnetic field of induction B has an energy ufieldassociated with it, given by

o

2

field 2 B

u = , (4.33)

where o is the permeability of free space. The totalenergy E field contained in the field is therefore given by

o

2

field 2 B

V E = , (4.34)

where V is the volume of the emitting region. Numerically, we have

( ) ( ) ( )23field kpcJ T BV TBD E = , (4.35)

The total energy E total in the lobes needed to give theobserved luminosity is therefore given, from equations(3.72) and (4.34), by

( ) .2

1o

22/1

o2/3

2/1

field particlestotal

BV LB

ba

K

E E E

+

+=

+

(4.36)

Unfortunately it is rarely possible to measure themagnetic field independently. How, therefore, are we to

estimate the total energy E total ? The first term inequation (4.36) is a decreasing function of the magneticinduction B whereas the second term is an increasingfunction of B. There must therefore be a minimum inthe total energy needed to give an observed luminosity

L. This is shown in Figure 4.7, which plots thelogarithm of the particle, field and total energies againstthe logarithm of the induction.

E field Eparticles

E total

log B

log E

Figure 4.7. Variation of particle (dashed curve), field (dotted curve) and total (solid curve)energies with B.

It is usually assumed that the magnetic field has thevalue that minimises the total energy for no better reason than that, even with this assumption, the totalenergies required of around 10 54 J areembarrassingly high!

To find the minimum, we differentiate E total with respectto B and set the result equal to zero to get Bmin . We find

( )7/2

oo

2/1

min 123

+

=

V

LK

ba

B

. (4.37)

The first term in round brackets depends solely onfundamental constants. The terms in square brackets areestimated or observed quantities.


( ) ( ) ( ) ( )( )7/2

3o

minm

HzW123.0T

+=

V

LK B

. (4.38)

It is easy to show, from equations (4.32) and (4.35) thatthe magnetic induction Bmin , makes the energy in the

particles nearly equal to that in the magnetic field:


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34

field

particles = E

E . (4.39)

Some people prefer to start with the assumption that theenergy is equally divided between the particles and thefield the so-called equipartition of energy rather thanto assume that the energy is minimised. From equation(4.39), it is obviously immaterial which assumption ismade.

From equations (3.52) and (3.75), we have for theminimum total energy E min ,

( ) .123

67

34

1

7/42/1

o

2/17/3

o

minfield,min

+

=

=

+=

Lb

aK

V

E E

(4.40)

4.2.2 THE SUPPLY OF E NERGY TO THE LOBES

Perhaps the most natural assumption, given their disposition on either side of the galaxy, is that the radiolobes were ejected as entities from the galaxy and have

been travelling independently since throughintergalactic space. There is certainly nothing in thevicinity of the lobes themselves that could beresponsible for them. The lobes are tens of kiloparsecs some 50 kpc or 150 000 light years in the case of Cygnus A, for example or more away from the galaxy.As we shall see later, they are moving out into theintergalactic medium at no more than a tenth of thevelocity of light so that they would have taken ~10 6 years to get there. Yet it can be shown from equation(4.25) that the lifetime of an electron in a radio lobe istypically ~10 5 years. The electrons would therefore havelost all their energy in the time taken by the lobes to getto their present position.

In fact, the problem is far worse than appears at firstsight! If the lobes had been ejected from the centralgalaxy, it is reasonable to assume that they had comefrom the nucleus of the galaxy where the activity isseen rather than from the benign elliptical galaxysurrounding the core. But the variability on the time-scale of a year or less of the activity in the nucleusshows that it must be less that a few parsecs in size. Letus take an upper limit of 10 parsecs. The radio lobesthemselves, however are of the order of severalkilo parsecs. Hence, the lobes must have undergone

expansion by a factor of around 1000 in their passagefrom the nucleus of the galaxy. Let us denote the factor of expansion by f . The we can say that the lobe expandsform an initial size d to a final size d' given by fd .Symbolically:

fd d d = . (4.41)

Now the lobes contain plasma highly ionised matter and the magnetic field B is therefore frozen into the

matter 12. As the lobes expand, therefore, the field linesget further apart and the field itself decreases.Quantitatively, the flux threading the lobe must beconserved. Since the size of the lobe is d , the flux threading it is given by

constant~ 22 == d B Bd , (4.42)

where B is the final value of the induction. Fromequations (4.41) and (4.42), therefore,

B f B 2= . (4.43)

The total energy E field within the field after expansion isgiven by

,2

2~

2

field1

o

2433

o

23

o

2

field

E f B f

d f

Bd

BV E

==

=

(4.44)

where V' is the expanded volume of the lobes and E field is the total field energy before expansion.

What about the electrons? The radius a of the orbit of ahighly-relativistic electron of energy E in the magneticfield B is given by

eBc E

a = (4.45)

so that the flux a threading the electron's orbit is given by

B E

ce Baa

2

222 1~ = . (4.46)

For changes that are slow compared with the time tocomplete one orbit that is, for adiabatic changes a must be constant so that

B E

B E 22 =

(4.47)

or, using equation (4.18),

E f E 1= . (4.48)

If no new electrons are added to the plasma, the totalnumber N of electrons contained in the lobes must beconstant so that

( )( ) .11

particles11

particles

E f E Nf K

E N K E =+=

+=(4.49)

12Strictly, the conductivity of the plasma would have to be infinite for the field to be completely tied to the matter, but the approximation isgood.


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The field and particle energies therefore scale in thesame way with the expansion factor f , preserving theequipartition between the particle and field energies:

Using equation (3.84) for the power radiated by anindividual electron, we have for the total power P total radiated by the lobes after expansion

,total62224

22total

P f E f B Nbf

E B NbP ==

= (4.50)

where P total is the total power radiated before theexpansion. Equation (4.50) says that the power radiated

by the lobes decreases by a factor f 6 during theexpansion. As we have seen, f is of order 10 3 so the

power decreases by eighteen orders of magnitude duringthe expansion! If the ejection hypothesis were true,therefore, one would expect to find that sources withsmall separation between the lobes and the centralgalaxy were on average more powerful than the larger sources. On the contrary, it is the larger sources that

are, on the whole, more powerful. We must thereforeseek another explanation of the origin of the radio lobes.

The accepted explanation is that there is a beam or jet of particles emanating from the central galaxy whichfeeds the lobes with fresh relativistic electrons. As wehave seen, there is direct evidence for such beams inradio sources.

4.2.3 MOTION OF THE JETS AND LOBES

4.2.3.1 Superluminal Velocity in Jets

Figure 4.8 shows knots or blobs of material in a jetleaving a central galaxy with velocity V . By taking radiomeasurements separated by months or years of these

blobs, we can measure their proper motion , that istheir angular velocity on the sky. If we can establish thedistance r o the galaxy, we can calculate the rate of change p&of the projected distance p :

or p =& . (4.51)

V

lp

ro

r

galaxy

Figure 4.8. Illustration of superluminal motion.

As we have already seen, p& is found to be greater thanthe velocity of light c in several sources. I shall nowshow that this an optical illusion and does not imply thatthe blobs are violating the tenets of relativity. Suppose

the furthest blob was ejected from the galaxy at time t e.The radiation emitted by the blob at this time is receivedat the Earth at time t r given by

co

er r

t t += , (4.52)

where r o is the distance of the galaxy from Earth13

. Atsome time t e + t e, the blob is in the position shown inthe diagram and is distant r o - r from the Earth. In this

position, it emits radiation which is received on Earth attime t r + t r where

( ) ( )c

r r t t t t

++=+ oeer r . (4.53)

From equations (4.52) and (4.53), we have

cr

t t = er . (4.54)

Note that the time between the reception of the twosignals is less than the time between emission becausethe light has less far to travel: it is this difference whichgives rise to the optical illusion.

In the time t r , the blob is observed to have moved adistance p perpendicular to the line-of-sight where,from Figure 4.8,

sinsin e == t V l p , (4.55)

where V is the velocity of the blob. From equations(4.54) and (4.55), we have

=

=

e

r

e

r 11

sinsin

t r

c

V V t

t

t p

p

& . (4.56)

But it is clear from the figure that

coscos e == t V lr (4.57)

so that

c

cV

V p

=

=

cos1

sin

cos1

sin& , (4.58)

where

cV =: (4.59)

is the velocity of the blob expressed as a fraction of thevelocity of light. Obviously, what we need if we are to

13For simplicity, I assume that the galaxy is at rest with respect to theEarth. This assumption does not affect the conclusion.


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observe a blob apparently moving as fast as, or faster than, light is to have

1cos1

sin

. (4.60)

We can re-arrange equation (4.60) to give

cossin1+

(4.61)

and it is easy to that the minimum value min of neededfor apparent superluminal velocity is 1/ 2:

cc

cV 707.02

minmin == . (4.62)

The fact that we do see such motion shows that thematerial in the jets feeding the radio lobes is movingrelativistically 14. Note that (sin +cos ) is symmetrical

about = /4 so that the same value of is needed for the angle ( /4 ) as for ( /4 + ). The reason for this is that, what we lose in a longer light travel time(cos decreasing) we gain in greater projected distance(sin increasing), and vice versa .

Inequality (4.61) can be manipulated to give the rangeof values for which will given superluminal velocitiesfor a given value of :

+

121

21

sin12121

sin 2121

(4.63)

4.2.3.2 Velocity of the LobesHow can we get an estimate of the velocities with whichthe lobes are separating from the central galaxy? Wecan use the symmetry of the apparent images to get bothupper and lower limits 15.

V

V

D

VG

Optical galaxy

Figure 4.9. Lobe and galaxy velocities.

Figure 4.9 shows schematically a central galaxy between the two radio lobes that are separated by a

14It is important to realise that we are here speaking of the bulk motionof the jets. The electrons within the jets may also moving with

pseudo-random relativistic velocities.15There are asymmetrical sources, such as head-tail sources, towhich these arguments clearly do not apply.

distance D . Observation puts limits on the ratio of thedistance the galaxy has moved away from the line

joining the lobes:

~< D, (4.64)

where is to be determined from observation. Supposethat the lobes are separating from the galaxy withvelocity V whilst the galaxy itself has a component of velocity V G at right angles to the line joining the lobes.Then

Vt D ~ , (4.65)

whilst

t V G~ , (4.66)

where t is the age of the lobes. From (4.64), (4.65) and(4.66), we have

g

~V

V > . (4.67)

Because we cannot measure the velocity the galaxydirectly, we have to apply (4.67) statistically. Typicalrandom velocities of galaxies are around 500 km s -1 andtypical values of are less than 0.01 kpc, giving a lower limit to values of V of some tens of thousands of kilometres a second or ~ 0.1 c.

v

v

l1

l2

r1

r2

r0

p1

p2

Figure 4.10. Asymmetry of radio sources

We can also get an upper limit on the velocity from theappearance of the source. Figure 4.10 shows lobes 1 and2 of a radio source, separated from the central galaxy by

distances l1 and l2 respectively. Their projected distanceson the sky, at right angles to the observer's line of sight,are p1 and p2 respectively. Let the respective distancesfrom earth of lobe 1, the central galaxy and lobe 2 be r 1,r 0 and r 2. I suppose that the separation of the two lobesfrom the central galaxy was zero at time t o.

The argument is similar to that used in the discussion of superluminal velocity. Consider radiation received atthe earth from all three components at time t r . Theradiation from the central galaxy was emitted at time t e given by


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c

r t t 0r e = . (4.68)

Similarly, the radiation from the two lobes was emittedat times t I given by

c

r t t i

i=

r . (4.69)

where i takes the values 1 or 2 for the two lobesrespectively. From equations (4.68) and (4.69) we have

cr r

t t ii+= 0e . (4.70)

But, from the figure,

cos0 ii lr r m= , (4.71)

where the upper sign refers to lobe 1 and the lower tolobe 2. From equations (4.70) and (4.71), we have

( )

cl

t

clr r

t t

i

ii

cos

cos

e

00e

=

+=m

(4.72)

Note that the radiation we receive at any time from lobe1 is emitted later than that from lobe 2 received at thesame time because lobe 1 is nearer to us than lobe 2.Since the lobes have been travelling with velocity V for a time ( t i t 0), we have for the distances li of the lobesfrom the central galaxy at the time they emitted theradiation that is received on earth at time t r ,

( ) ( )

==c

lt t V t t V l iii

cos0e0 (4.73)

so that

( )

= cos1

0e

cV

t t V l i

m

(4.74)

and

cos1

cos1

21

cV c

V

ll

+= . (4.75)

Finally, the projected separations pi are given by

sinii l p = (4.76)

so that they are also related by the right hand side of equation (4.75):

cos1

cos1

2

1

cV cV

p

p

+= . (4.77)

Note that p1 is greater than p2 because the light startedlater from lobe 1, which therefore had a longer time totravel. If observation shows that

2

1

p p

, (4.78)

then equation (4.77)shows that

cV 11

cos +

. (4.79)

Unfortunately, the angle cannot be measured. We canapply inequality (4.79) statistically , though, assumingthat the orientation of radio galaxies is random. The

result is that the velocities of radio lobes do not exceedabout a tenth of the velocity of light.

4.2.4 CONFINEMENT OF THE RADIO LOBES

4.2.4.1 Diameter-Separation Ratio

Dd

V

V

Figure 4.11. Diameter-separation ratio of radio lobes.

Figure 4.11 is a sketch of a typical pair of radio lobes.Consider the ratio of the separation D of the lobes totheir diameter d :

d D=: (4.80)

Observed values of are around ten or more. Let usassume that the two lobes, constantly replenished withfresh electrons, have been moving way from the centralgalaxy with velocity V for a time t . Clearly,

ct D 2

, (4.81)

since otherwise they would have been travelling faster than light. Now, if there is nothing to stop them, thelobes themselves will expand and they will do so at the


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velocity of sound uS in the gas of which they arecomposed, giving

t ud S ~ . (4.82)

In general, the velocity of sound of sound in a fluid isgiven by (cf. Note 3)

S S

pu

=

2 (4.83)

where p is the pressure and the density. For arelativistic fluid, where the pressure is related to theenergy-density u by

2

31

31

cu p == , (4.84)

we have

3

cu S = . (4.85)

From relations (4.81), (4.82) and (4.85), we can deducethat

32~> m, equations (4.101) reduce to

.

;2

V V

V vv

+

(4.102)

In overtaking collisions, where v is in the same directionas V ,

V vv 2= , (4.103)

representing a loss in energy. In head-on collisions, onthe other hand, where v is in the opposite direction to V ,

V vv 2+= , (4.104)

so that the particle gains energy. If E and E' are theinitial and final energies respectively of the particle,then

( ) .221

21

21

;21

222

2

V vmvmvm E

mv E

===

=(4.105)

If v >> V , then

( )vV vm E 421 2 (4.106)

so that the change in energy E is given by

vV

E vV m E E E 4421 = . (4.107)

The change in the particles energy is therefore proportional to the energy itself

This is solution to the one-dimensional problem. For two- or three-dimensional scattering, the final velocitiesare under-determined unless we specify the anglethrough which the particle is scattered. But, for isotropicscattering , the one-dimensional solution is a goodapproximation to the average change in velocity.

Realistic scattering, will not necessarily be either isotropic or elastic. A more detailed analysis typicallygives a result similar to (4.107) but, with the factor of four replaced by unity:

.vV

E E . (4.108)

Assuming that the velocity of the particle is alreadycomparable to c, we have

.cV

E E . (4.109)

where the positive sign holds for head-on collisions andthe negative for overtaking collisions.

4.3.2.3 Growth of Energy

At first sight, we seem to have achieved nothing: whatwe gain in head-on collisions, we lose in overtakingones. But, head-on collisions are slightly more frequent

because the number of collisions per unit time dependson relative velocity vrel of particle and target (cf.

Note 3), the rate being given by

relvn R = , (4.110)

where n is the number-density of the scatterers and istheir cross-section. The rate R+ of head-on collisions istherefore given by

( )V cn R +=+ 21

, (4.111)

where the factor of one half reflects the fact that the particles are equally likely to be going in either direction. Similarly, the rate R- of overtaking collisionsis given by

( )V cn R = 21

. (4.112)

The total rate R is given by

cn R R R =+ + (4.113)

and the net rate Rnet by

V n R R R = + (4.114)

The net rate of energy gain E & is given by

( )

,2scatterer 2

cE nc

V E n

cV

E V ncV

E R R

E R E R E

==

==

+=

+

++&

(4.115)

where scatterer is the velocity of the scatterer measured asa fraction of the velocity of light:

cV =:scatterer . (4.116)

The average energy gain per collision E net is thereforegiven by


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( )

.2scatterer

2

net

=

=

E

cnc

V E n

R E

E &

(4.117)

Suppose the average time between collisions is . Eachcollision gives E net so that the energy gained dE intime dt is given by

net E dt

dE =

(4.118)

or

E E

dt dE == 2scatterer net . (4.119)

Equation (4.83) predicts exponential growth of energywith time:

=

t

E E 2scatterer o exp . (4.120)

Inversely, the time t ( E ) required to reach energy E isgiven by

( )

=

o2scatterer

ln1

E E

E t . (4.121)

4.3.2.4 Loss-rate

Let N ( E )dE be number density of particles with energiesin range E to E + dE . Suppose that the probability P of

an electrons escape from the scattering region isindependent of time and energy. Then, in time dt , thedecrease dN escape ( E ) in the density of particles throughescape is given by

( ) ( )Pdt E N E dN =escape (4.122)

so that the rate ( ) E N escaope& of escape of particles withenergy E is given by

( )( ) ( )

T E N

dt

E dN E N = escapeescaope& , (4.123)

where the leakage time T is given by

PT

1:= . (4.124)

4.3.2.5 Resultant Energy Distribution

The equation of continuity equation in one-dimensionalenergy-space is 18

18 This can be obtained by adapting the continuity equation of Note 3to one dimension.

( ) ( ) ( ) E N dt dE

E N E t

E N escape

&=

+

. (4.125)

Let us consider the steady state in which the density N ( E ) of electrons in energy-space is constant.Setting ( ) t E N to zero in equation (4.125) andsubstituting from equation (4.119) for dE /dt , we have

( ) ( )T E N E

E N E

=

2scatterer , (4.126)

which has solution

( )

=

oo E

E N E N , (4.127)

where N o and E o are constants and

T

2

scatterer

11 += . (4.128)

Hence this so-called second-order Fermi acceleration mechanism gives the desired power-law. Unfortunately, , and scatterer depend upon the physical processesgoing on in the source and does not predict theuniversal value of 2.5 for . Let us try somethingdifferent.

4.3.3 SHOCK ACCELERATION

4.3.3.1 Energy-Gain

VSVf

Figure 4.14. Shock acceleration.

An alternative mechanism is provided by the sort of shock fronts that we might expect to exist in the highlyexcited regions that exist around AGN. Figure 4.14shows in the rest frame of the shock front scatteringcentres both upstream and downstream of the shock.These centres are more or less at rest in the materialsurrounding them and are therefore approaching eachother at velocity V , the difference in velocity betweenthe speed relative to the shock of the upstream anddownstream material:

f S V V V = , (4.129)


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where V S is the velocity of the shock advancing into theunshocked region and V f is velocity of the materialflowing away downstream of the shock front (cf.

Note 3).

l

Line of upstream scattersLine of downstream scatters

Shock front c

Figure 4.15. Multiple crossing of shock front.

Now consider electrons moving, at velocities nearlyequal to c, crossing the shock front, bouncing off ascatterer, travelling back across the front, bouncing off another scatterer to re-cross the front again, etc, asshown schematically in Figure 4.15. The electrons meetthe scatterers head on each time so that the gain inenergy at each double crossing of the front is given bytwice equation (4.108) with the positive sign and with V replaced by V :

cV

E E = 2 . (4.130)

Compare this first-order process, with the random

collisions of section 4.3.2, in which the fractionalchange in energy was only second-order in V /c. In the present case, the equivalent of equation (4.119) is

E E

dt dE == , (4.131)

where is the mean time between scatterings, as before,and where is now given by

cV =2 . (4.132)

4.3.3.2 Loss-rate

What about the time T required for the loss of electronsfrom the scattering region? Although the electrons areindividually moving almost with velocity c, the net rateat which material crosses the front is only V , thedifference between the upstream and downstreamvelocities. If l is the typical distance between scattererson either side of the front, we must have

V T

~l

. (4.133)

On the other hand, the electron crosses the front manytimes during this process at velocity c so that the meantime between double scatterings, is given by

c~2

l. (4.134)

From equations (4.133), (4.132) and (4.134), therefore,

=

cc

V T ~~

l. (4.135)

This is the key to getting a process-independent spectralindex: the escape time is directly related to the meantime between scatterings.

4.3.3.3 Resultant Energy Distribution

With this new mechanism, the solution to equation(4.126) is still given by (4.127):

( )

= oo E E

N E N

but is now given by

( )2

11

11 =+=+=

T , (4.136)

where I have used equation (4.135). We were trying toget a value of 2.5 for and have managed to get a valueof 2. Not quite, therefore, but pretty close and at leasttotally independent of the detailed conditions in thesource. It is likely, therefore that some modification of the process, including the effects of radiation losses,

will prove to be the correct model.

Bibliography for Chapter 4

[1] Robson, I. Active Galactic Nuclei , Wiley 1996.ISBN 0-471-96050-0

[2] Field, G B, Arp, H and Bahcall, J N, The Redshift Controversy , W A Benjamin Inc, MA, 1973

[3] Shu, F H. The Physical Universe , UniversityScience Books, CA, 1982. ISBN 0-19-855-706-X

Documents

ASTM052 Extragalactic Astrophysics Notes 4 of 6 (QMUL)